For every linear map $M: V \to W$ between real finite dimensional vector spaces one can always choose a basis of $W$ and one of $V$ so that the matrix representation of $M$ has $n :=$rank$(M)$ ones along the diagonal and only zeros elsewhere. Now let $V = W$, we can compute its eigenvalues and the standard form of its matrix representation would contain only non-zero eigenvalues of the diagonal. Why can we not change the basis to make all the entries equal to one, as before?
I thought that if we choose a basis of eigenvectors, we would have have the eigenvalues of the diagonal (as i.e. in Jordan normal form). Now, if we scale those eigenvectors with their eigenvalues, we should have ones along the diagonal, but why isn't that possible?
Because now we only have one basis to deal with, whereas when we are dealing with two vector spaces we can deal with two bases, one for each space.